## Trigonometry (11th Edition) Clone

$$\sin165^\circ=\frac{\sqrt{2-\sqrt3}}{2}$$
$$\sin165^\circ$$ We would write $165^\circ$ as the difference of $180^\circ$ and $15^\circ$, as the angle $15^\circ$ can be applied the half-angle identities. $$\sin165^\circ=\sin(180^\circ-15^\circ)$$ - Difference Identity for sine: $\sin(A+B)=\sin A\cos B+\cos A\sin B$ with $A=180^\circ$ and $B=15^\circ$ $$\sin165^\circ=\sin180^\circ\cos15^\circ-\cos180^\circ\sin15^\circ$$ $\cos180^\circ=-1$ and $\sin180^\circ=0$ $$\sin165^\circ=0\times\cos15^\circ-(-1)\times\sin15^\circ$$ $$\sin165^\circ=\sin15^\circ$$ *Calculate $\sin15^\circ$ $$\sin15^\circ=\sin\Big(\frac{30^\circ}{2}\Big)$$ Apply the identity $\sin\Big(\frac{A}{2}\Big)=\pm\sqrt{\frac{1-\cos A}{2}}$ with $A=30^\circ$. $$\sin15^\circ=\pm\sqrt{\frac{1-\cos30^\circ}{2}}$$ $$\sin15^\circ=\pm\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}$$ $$\sin15^\circ=\pm\sqrt{\frac{\frac{2-\sqrt3}{2}}{2}}$$ $$\sin15^\circ=\pm\sqrt{\frac{2-\sqrt3}{4}}$$ $$\sin15^\circ=\frac{\pm\sqrt{2-\sqrt3}}{2}$$ As $15^\circ$ is in quadrant I, where $\sin\theta\gt0$, so $\sin15^\circ\gt0$, so we would select the positive square root. $$\sin15^\circ=\frac{\sqrt{2-\sqrt3}}{2}$$ Therefore, $$\sin165^\circ=\sin15^\circ=\frac{\sqrt{2-\sqrt3}}{2}$$